Pure Si at 300K has equal electron `(n_(e))` and hole `(n_(h))` concentrations of `1.5xx10^(16)m^(-3)` Doping by indium increases `n_(h)` to `3xx10^(22)m^(-3)`. Calculate `n_(e)` in the doped Si.

Explanation

For a doped semi-conductor in thermal equilibrium `n_(e)n_(h) = n_(1)^(2)` (Law of mass action)
`n_(e)=(n_(1)^(2))/(h_(h))=((1.5xx10^(16))^(2))/(3xx10^(22))=7.5 xx10^(9)m^(-3)`